Samacheer Kalvi Class 12 Maths Solutions Chapter 3 Theory of Equations Exercise 3.5

 

Question 1. Solve the following equations
(i) \( \sin^2 x - 5 \sin x + 4 = 0 \)

Answer:
We have the equation: \( \sin^2 x - 5 \sin x + 4 = 0 \)
Let \( y = \sin x \). The equation becomes a quadratic in \( y \):
\( y^2 - 5y + 4 = 0 \)
Factor this quadratic:
\( (y - 1)(y - 4) = 0 \)
This gives two possible values for \( y \):
\( y = 1 \) or \( y = 4 \)
Substitute back \( y = \sin x \):
\( \sin x = 1 \) or \( \sin x = 4 \)
We know that the range of the sine function is \( -1 \le \sin x \le 1 \). So, \( \sin x = 4 \) is not possible.
We consider only \( \sin x = 1 \).
The general solution for \( \sin x = \sin \alpha \) is \( x = n\pi + (-1)^n \alpha \), where \( n \in \mathbb{Z} \).
For \( \sin x = 1 \), we have \( \sin x = \sin \frac{\pi}{2} \).
So, \( \alpha = \frac{\pi}{2} \).
Therefore, the solution for \( x \) is \( n\pi + (-1)^n \frac{\pi}{2} \).
In simple words: First, treat the sine term as a simple variable to solve a basic quadratic equation. Then, check which solutions for sine are actually possible, since sine values must be between -1 and 1. Finally, use the general formula for sine equations to find the values of x.

🎯 Exam Tip: Always remember the range of trigonometric functions (like \( \sin x \) and \( \cos x \) between -1 and 1) to eliminate impossible solutions in such equations.

 

Question 1.
(ii) \( 12x^3 + 8x = 29x^2 - 4 = 0 \)

Answer:
First, rearrange the equation into standard polynomial form, setting it to zero:
\( 12x^3 - 29x^2 + 8x + 4 = 0 \)
We can test for simple rational roots. Let's try \( x = 2 \):
\( 12(2)^3 - 29(2)^2 + 8(2) + 4 = 12(8) - 29(4) + 16 + 4 \)
\( = 96 - 116 + 16 + 4 = 116 - 116 = 0 \)
Since the result is 0, \( x = 2 \) is a root. We can use synthetic division to find the remaining quadratic factor:

The quotient polynomial is \( 12x^2 - 5x - 2 = 0 \).
Now, we factor this quadratic equation:
\( 12x^2 - 5x - 2 = 0 \)
We can split the middle term \( -5x \) into \( -8x + 3x \):
\( 12x^2 - 8x + 3x - 2 = 0 \)
Factor by grouping:
\( 4x(3x - 2) + 1(3x - 2) = 0 \)
\( (3x - 2)(4x + 1) = 0 \)
This gives two more roots:
\( 3x - 2 = 0 \implies 3x = 2 \implies x = \frac{2}{3} \)
\( 4x + 1 = 0 \implies 4x = -1 \implies x = -\frac{1}{4} \)
Therefore, the roots of the equation are \( 2, \frac{2}{3}, \text{ and } -\frac{1}{4} \).
In simple words: First, put the equation in order. Try to find a simple whole number that makes the equation zero (like x=2). Once you find one, use division to get a simpler quadratic equation. Then, solve that quadratic equation to find the other two solutions.

🎯 Exam Tip: When testing for integer roots, start with small integers like \( \pm 1, \pm 2, \pm 3 \). For a cubic equation, finding one root reduces it to a quadratic, which can then be solved using factoring or the quadratic formula.

 

Question 2. Examine for the rational roots of
(i) \( 2x^3 - x^2 - 1 = 0 \)

Answer:
We are given the polynomial equation \( 2x^3 - x^2 - 1 = 0 \).
A quick way to check if \( x = 1 \) is a root is to sum the coefficients:
Sum of coefficients \( = 2 - 1 + 0 - 1 = 0 \).
Since the sum of coefficients is 0, \( x = 1 \) is a root of the equation.
Now, we use synthetic division with \( x = 1 \) to find the remaining quadratic factor:

The quotient polynomial is \( 2x^2 + x + 1 = 0 \).
To find the roots of this quadratic, we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = 1, c = 1 \).
\( x = \frac{-1 \pm \sqrt{1^2 - 4(2)(1)}}{2(2)} \)
\( x = \frac{-1 \pm \sqrt{1 - 8}}{4} \)
\( x = \frac{-1 \pm \sqrt{-7}}{4} \)
\( x = \frac{-1 \pm i\sqrt{7}}{4} \)
Since these roots involve \( i \) (the imaginary unit), they are imaginary roots and not rational.
Therefore, the only rational root of the equation is \( x = 1 \).
In simple words: To find rational roots, first check if \( x=1 \) is a root by adding up all the numbers in front of the variables. If it is, divide the equation by \( (x-1) \) to get a simpler equation. Then, use the quadratic formula on the remaining part to see if there are any more easy-to-find number solutions. If the answer involves `i`, they are not rational.

🎯 Exam Tip: For polynomial equations, the Rational Root Theorem states that any rational root \( p/q \) must have \( p \) as a factor of the constant term and \( q \) as a factor of the leading coefficient. In this case, \( p \) divides -1 (so \( \pm 1 \)) and \( q \) divides 2 (so \( \pm 1, \pm 2 \)). Possible rational roots are \( \pm 1, \pm 1/2 \).

 

Question 2.
(ii) \( x^8 - 3x + 1 = 0 \)

Answer:
For the polynomial equation \( x^8 - 3x + 1 = 0 \):
The leading coefficient (coefficient of \( x^8 \)) is \( a_n = 1 \).
The constant term is \( a_0 = 1 \).
According to the Rational Root Theorem, any possible rational root \( \frac{p}{q} \) must have \( p \) as a divisor of the constant term \( a_0 \) and \( q \) as a divisor of the leading coefficient \( a_n \).
Here, \( p \) must divide \( 1 \), so \( p = \pm 1 \).
And \( q \) must divide \( 1 \), so \( q = \pm 1 \).
Therefore, the only possible rational roots are \( \pm \frac{1}{1} \), which means \( x = 1 \) or \( x = -1 \).
Let's test these values:
For \( x = 1 \):
\( (1)^8 - 3(1) + 1 = 1 - 3 + 1 = -1 \)
Since \( -1 \neq 0 \), \( x = 1 \) is not a root.
For \( x = -1 \):
\( (-1)^8 - 3(-1) + 1 = 1 + 3 + 1 = 5 \)
Since \( 5 \neq 0 \), \( x = -1 \) is not a root.
As neither of the possible rational roots actually satisfy the equation, we can conclude that the equation \( x^8 - 3x + 1 = 0 \) has no rational roots.
In simple words: First, find the number in front of the biggest power of x and the number without any x. Then, find all numbers that can divide these. These are the only possible easy-to-find fraction answers. Try putting each of these numbers into the equation. If none of them work, then there are no simple fraction solutions.

🎯 Exam Tip: The Rational Root Theorem is a powerful tool to narrow down potential rational roots. Always test all candidates systematically.

 

Question 3. Solve: \( 8x^{3/2n} - 8x^{-3/2n} = 63. \)

Answer:
We have the equation \( 8x^{3/2n} - 8x^{-3/2n} = 63 \).
Let \( y = x^{3/2n} \). Then \( x^{-3/2n} = \frac{1}{x^{3/2n}} = \frac{1}{y} \).
Substitute \( y \) into the equation:
\( 8y - \frac{8}{y} = 63 \)
Multiply the entire equation by \( y \) to clear the denominator (assuming \( y \neq 0 \)):
\( 8y^2 - 8 = 63y \)
Rearrange this into a standard quadratic equation:
\( 8y^2 - 63y - 8 = 0 \)
We can factor this quadratic equation by splitting the middle term. We need two numbers that multiply to \( 8 \times (-8) = -64 \) and add up to \( -63 \). These numbers are \( -64 \) and \( 1 \).
\( 8y^2 - 64y + y - 8 = 0 \)
Factor by grouping:
\( 8y(y - 8) + 1(y - 8) = 0 \)
\( (y - 8)(8y + 1) = 0 \)
This gives two possible values for \( y \):
\( y - 8 = 0 \implies y = 8 \)
\( 8y + 1 = 0 \implies y = -\frac{1}{8} \)
Now, we substitute back \( y = x^{3/2n} \):
Case 1: \( y = 8 \)
\( x^{3/2n} = 8 \)
Since \( 8 = 2^3 \), we have \( x^{3/2n} = 2^3 \)
Raise both sides to the power of \( \frac{2n}{3} \) to solve for \( x \):
\( (x^{3/2n})^{2n/3} = (2^3)^{2n/3} \)
\( x = 2^{(3 \cdot 2n)/3} \)
\( x = 2^{2n} \)
\( x = (2^2)^n \)
\( x = 4^n \)
Case 2: \( y = -\frac{1}{8} \)
\( x^{3/2n} = -\frac{1}{8} \)
For \( x^{3/2n} \) to be a real number, if \( 2n \) is an even integer (which it is if \( n \) is an integer), then \( x \) must be non-negative for the expression \( x^{1/2n} \) to be real and defined in the usual sense. If \( x \ge 0 \), then \( x^{3/2n} \) must also be non-negative. Since \( -\frac{1}{8} \) is negative, this case yields no real solutions for \( x \).
Therefore, the only real root for the equation is \( x = 4^n \).
In simple words: This equation looks complicated because of the fractional power. Make it simpler by replacing the repeating part (like `x` to some power) with a new letter, say `y`. Solve for `y` using normal algebra. Once you find `y`, put the original `x` expression back in and solve for `x`. Sometimes one of the `y` answers might not work for `x` if it leads to impossible math, like taking an even root of a negative number.

🎯 Exam Tip: When dealing with variables raised to fractional powers like \( x^{m/n} \), be careful with negative results. If \( n \) is even, \( x^{1/n} \) is only defined for non-negative \( x \) in real numbers, and its value is taken as positive. If \( n \) is odd, \( x^{1/n} \) is defined for all real \( x \).

 

Question 4. Solve: \( 2\sqrt{\frac{x}{a}} + 3\sqrt{\frac{a}{x}} = \frac{b}{a} = \frac{6a}{b} \)

Answer:
The question implies that \( \frac{b}{a} = \frac{6a}{b} \), which gives \( b^2 = 6a^2 \). This provides a relationship between \( a \) and \( b \). The equation to solve for \( x \) is \( 2\sqrt{\frac{x}{a}} + 3\sqrt{\frac{a}{x}} = \frac{b}{a} + \frac{6a}{b} \).
Let \( y = \sqrt{\frac{x}{a}} \). Then, \( \sqrt{\frac{a}{x}} = \frac{1}{\sqrt{x/a}} = \frac{1}{y} \).
Substitute \( y \) into the equation:
\( 2y + \frac{3}{y} = \frac{b}{a} + \frac{6a}{b} \)
Square both sides of the equation:
\( \left(2y + \frac{3}{y}\right)^2 = \left(\frac{b}{a} + \frac{6a}{b}\right)^2 \)
Expand both sides:
\( (2y)^2 + 2(2y)\left(\frac{3}{y}\right) + \left(\frac{3}{y}\right)^2 = \left(\frac{b}{a}\right)^2 + 2\left(\frac{b}{a}\right)\left(\frac{6a}{b}\right) + \left(\frac{6a}{b}\right)^2 \)
\( 4y^2 + 12 + \frac{9}{y^2} = \frac{b^2}{a^2} + 12 + \frac{36a^2}{b^2} \)
Subtract 12 from both sides:
\( 4y^2 + \frac{9}{y^2} = \frac{b^2}{a^2} + \frac{36a^2}{b^2} \)
Now, substitute back \( y^2 = \frac{x}{a} \):
\( 4\left(\frac{x}{a}\right) + \frac{9}{\left(\frac{x}{a}\right)} = \frac{b^2}{a^2} + \frac{36a^2}{b^2} \)
\( \frac{4x}{a} + \frac{9a}{x} = \frac{b^2}{a^2} + \frac{36a^2}{b^2} \)
This equation can be rearranged and factored to find the values of \( x \). The roots are determined to be:
\( x = \frac{b^2}{4a} \) or \( x = \frac{9a^3}{b^2} \)
In simple words: First, simplify the complex square root parts by using a temporary letter `y`. Then, square both sides of the equation to get rid of the roots. Replace `y` with its original expression involving `x` to get an equation with `x`. Solve this final equation for `x` to find the answers.

🎯 Exam Tip: When dealing with equations involving square roots of reciprocal terms, substitution is often the best approach. Squaring both sides is a common technique, but be careful of introducing extraneous roots, which would need to be checked against the original equation.

 

Question 5.
(i) \( 6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0 \)

Answer:
This is a Type I even-degree reciprocal equation because the coefficients are symmetric (`6, -35, 62, -35, 6`).
Since \( x=0 \) is not a root (substituting \( x=0 \) gives \( 6 \neq 0 \)), we can divide the entire equation by \( x^2 \):
\( \frac{6x^4}{x^2} - \frac{35x^3}{x^2} + \frac{62x^2}{x^2} - \frac{35x}{x^2} + \frac{6}{x^2} = 0 \)
\( 6x^2 - 35x + 62 - \frac{35}{x} + \frac{6}{x^2} = 0 \)
Group the terms with similar powers:
\( 6\left(x^2 + \frac{1}{x^2}\right) - 35\left(x + \frac{1}{x}\right) + 62 = 0 \)
Let \( y = x + \frac{1}{x} \).
Squaring both sides gives \( y^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^2} = x^2 + 2 + \frac{1}{x^2} \).
So, \( x^2 + \frac{1}{x^2} = y^2 - 2 \).
Substitute these into the equation:
\( 6(y^2 - 2) - 35y + 62 = 0 \)
\( 6y^2 - 12 - 35y + 62 = 0 \)
\( 6y^2 - 35y + 50 = 0 \)
Now, we factor this quadratic equation for \( y \). We look for two numbers that multiply to \( 6 \times 50 = 300 \) and add up to \( -35 \). These numbers are \( -20 \) and \( -15 \).
\( 6y^2 - 20y - 15y + 50 = 0 \)
Factor by grouping:
\( 2y(3y - 10) - 5(3y - 10) = 0 \)
\( (3y - 10)(2y - 5) = 0 \)
This gives two values for \( y \):
\( 3y - 10 = 0 \implies y = \frac{10}{3} \)
\( 2y - 5 = 0 \implies y = \frac{5}{2} \)
Now, substitute back \( y = x + \frac{1}{x} \) for each case:
Case 1: \( x + \frac{1}{x} = \frac{10}{3} \)
Multiply by \( 3x \): \( 3x^2 + 3 = 10x \)
Rearrange: \( 3x^2 - 10x + 3 = 0 \)
Factor: \( (3x - 1)(x - 3) = 0 \)
So, \( 3x - 1 = 0 \implies x = \frac{1}{3} \) and \( x - 3 = 0 \implies x = 3 \).
Case 2: \( x + \frac{1}{x} = \frac{5}{2} \)
Multiply by \( 2x \): \( 2x^2 + 2 = 5x \)
Rearrange: \( 2x^2 - 5x + 2 = 0 \)
Factor: \( (2x - 1)(x - 2) = 0 \)
So, \( 2x - 1 = 0 \implies x = \frac{1}{2} \) and \( x - 2 = 0 \implies x = 2 \).
Thus, the roots of the equation are \( \frac{1}{3}, 3, \frac{1}{2}, \text{ and } 2 \).
In simple words: For this type of equation where numbers are the same at the start and end, divide by `x` squared. Then, replace `x + 1/x` with `y` to make it a simpler equation. Solve for `y`, and then put `x + 1/x` back in to find the four values of `x`. This special trick helps solve these longer equations more easily.

🎯 Exam Tip: Reciprocal equations are common. Always remember the substitution \( y = x + \frac{1}{x} \) and that \( x^2 + \frac{1}{x^2} = y^2 - 2 \). This simplifies the quartic equation into a quadratic.

 

Question 5.
(ii) \( x^4 + 3x^3 - 3x - 1 = 0 \)

Answer:
We have the polynomial equation \( x^4 + 3x^3 - 3x - 1 = 0 \).
Let's test for simple integer roots.
For \( x = 1 \):
\( (1)^4 + 3(1)^3 - 3(1) - 1 = 1 + 3 - 3 - 1 = 0 \). So, \( (x - 1) \) is a factor.
For \( x = -1 \):
\( (-1)^4 + 3(-1)^3 - 3(-1) - 1 = 1 + 3(-1) + 3 - 1 = 1 - 3 + 3 - 1 = 0 \). So, \( (x + 1) \) is also a factor.
Since both \( (x - 1) \) and \( (x + 1) \) are factors, their product \( (x - 1)(x + 1) = x^2 - 1 \) is also a factor.
We can divide the original polynomial by \( (x^2 - 1) \). First, we can use synthetic division twice, once for \( x=1 \) and then for \( x=-1 \).
Using \( x=1 \):

The resulting cubic polynomial is \( x^3 + 4x^2 + 4x + 1 \).
Now, use synthetic division with \( x=-1 \) on this cubic polynomial:
The final quadratic factor is \( x^2 + 3x + 1 = 0 \).
So, the original equation can be factored as:
\( (x - 1)(x + 1)(x^2 + 3x + 1) = 0 \)
Now, we find the roots from each factor:
\( x - 1 = 0 \implies x = 1 \)
\( x + 1 = 0 \implies x = -1 \)
For \( x^2 + 3x + 1 = 0 \), we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( x = \frac{-3 \pm \sqrt{3^2 - 4(1)(1)}}{2(1)} \)
\( x = \frac{-3 \pm \sqrt{9 - 4}}{2} \)
\( x = \frac{-3 \pm \sqrt{5}}{2} \)
Therefore, the roots of the equation are \( 1, -1, \frac{-3 + \sqrt{5}}{2}, \text{ and } \frac{-3 - \sqrt{5}}{2} \).
In simple words: Start by testing if `x = 1` or `x = -1` are solutions, which is a common trick for polynomial equations. If they are, you can divide the original equation by `(x-1)` and `(x+1)` (or `x^2 - 1`) to get a simpler quadratic equation. Finally, use the quadratic formula to solve this simpler equation for the remaining solutions.

🎯 Exam Tip: Recognizing factors like \( (x-1) \) or \( (x+1) \) early can significantly simplify solving higher-degree polynomial equations. Always perform synthetic division carefully, especially with placeholder zeros for missing terms.

 

Question 6. Find all real numbers satisfying
\( 4^x - 3 (2^{x+2}) + 2^5 = 0 \)

Answer:
We are given the equation \( 4^x - 3 (2^{x+2}) + 2^5 = 0 \).
First, rewrite the terms to have a common base, which is 2:
\( (2^2)^x - 3 (2^x \cdot 2^2) + 32 = 0 \)
\( (2^x)^2 - 3 \cdot 2^x \cdot 4 + 32 = 0 \)
Simplify the coefficients:
\( (2^x)^2 - 12 \cdot 2^x + 32 = 0 \)
Now, let \( y = 2^x \). Substitute \( y \) into the equation:
\( y^2 - 12y + 32 = 0 \)
This is a standard quadratic equation. Factor it to find the values of \( y \). We need two numbers that multiply to 32 and add up to -12. These are -4 and -8.
\( (y - 4)(y - 8) = 0 \)
This gives two possible values for \( y \):
\( y - 4 = 0 \implies y = 4 \)
\( y - 8 = 0 \implies y = 8 \)
Now, substitute back \( y = 2^x \) for each case:
Case 1: \( y = 4 \)
\( 2^x = 4 \)
Since \( 4 = 2^2 \), we have \( 2^x = 2^2 \)
Comparing the exponents, we get \( x = 2 \).
Case 2: \( y = 8 \)
\( 2^x = 8 \)
Since \( 8 = 2^3 \), we have \( 2^x = 2^3 \)
Comparing the exponents, we get \( x = 3 \).
Thus, the real numbers satisfying the equation are \( 2 \) and \( 3 \).
In simple words: Change all numbers in the equation to have the same base (like 2 in this case). Then, replace the repeated `base` to the power of `x` with a new letter, making it a simple quadratic equation. Solve for this new letter, and then put the `base` to the power of `x` back to find the actual `x` values. This method simplifies complex exponential equations.

🎯 Exam Tip: When solving exponential equations, always try to express all terms with the same base. This allows you to convert the equation into a simpler algebraic form by comparing exponents or using substitution.

 

Question 7. Solve the equation \( 6x^4 - 5x^3 - 38x^2 - 5x + 6 = 0 \) if it is known that \( \frac{1}{3} \) is a solution.

Answer:
The given equation is \( 6x^4 - 5x^3 - 38x^2 - 5x + 6 = 0 \).
This is a reciprocal equation of even degree because the coefficients are symmetric (`6, -5, -38, -5, 6`).
A key property of reciprocal equations is that if \( \alpha \) is a root, then \( \frac{1}{\alpha} \) is also a root.
We are given that \( x = \frac{1}{3} \) is a solution. Therefore, \( x = \frac{1}{1/3} = 3 \) must also be a solution.
Since \( x=0 \) is not a root, we can divide the entire equation by \( x^2 \):
\( 6x^2 - 5x - 38 - \frac{5}{x} + \frac{6}{x^2} = 0 \)
Group terms with similar patterns:
\( 6\left(x^2 + \frac{1}{x^2}\right) - 5\left(x + \frac{1}{x}\right) - 38 = 0 \)
Let \( y = x + \frac{1}{x} \).
Then, \( y^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^2} = x^2 + 2 + \frac{1}{x^2} \).
So, \( x^2 + \frac{1}{x^2} = y^2 - 2 \).
Substitute these into the equation:
\( 6(y^2 - 2) - 5y - 38 = 0 \)
\( 6y^2 - 12 - 5y - 38 = 0 \)
\( 6y^2 - 5y - 50 = 0 \)
Now, we factor this quadratic equation for \( y \). We need two numbers that multiply to \( 6 \times (-50) = -300 \) and add up to \( -5 \). These numbers are \( -20 \) and \( 15 \).
\( 6y^2 - 20y + 15y - 50 = 0 \)
Factor by grouping:
\( 2y(3y - 10) + 5(3y - 10) = 0 \)
\( (3y - 10)(2y + 5) = 0 \)
This gives two values for \( y \):
\( 3y - 10 = 0 \implies y = \frac{10}{3} \)
\( 2y + 5 = 0 \implies y = -\frac{5}{2} \)
Now, substitute back \( y = x + \frac{1}{x} \) for each case:
Case 1: \( x + \frac{1}{x} = \frac{10}{3} \)
Multiply by \( 3x \): \( 3x^2 + 3 = 10x \)
Rearrange: \( 3x^2 - 10x + 3 = 0 \)
Factor: \( (3x - 1)(x - 3) = 0 \)
So, \( 3x - 1 = 0 \implies x = \frac{1}{3} \) and \( x - 3 = 0 \implies x = 3 \).
Case 2: \( x + \frac{1}{x} = -\frac{5}{2} \)
Multiply by \( 2x \): \( 2x^2 + 2 = -5x \)
Rearrange: \( 2x^2 + 5x + 2 = 0 \)
Factor: \( (2x + 1)(x + 2) = 0 \)
So, \( 2x + 1 = 0 \implies x = -\frac{1}{2} \) and \( x + 2 = 0 \implies x = -2 \).
Thus, the four roots of the equation are \( \frac{1}{3}, 3, -\frac{1}{2}, \text{ and } -2 \).
In simple words: This is a special type of equation where if one solution is `1/3`, then `3` is also a solution. You can simplify the equation by dividing by `x` squared and then replacing `x + 1/x` with `y`. Solve for `y`, and then substitute back to find all four `x` values.

🎯 Exam Tip: Recognizing a reciprocal equation (coefficients symmetric) and utilizing the property that if \( \alpha \) is a root, \( 1/\alpha \) is also a root, simplifies the problem significantly, especially when one root is provided.